Let $X_t$ be a real-valued stochastic process (if it helps, we can assume that it is a component of a multivariate diffusion or jump-diffusion process). I'm looking for sufficient conditions under which $X_t$ diverges in probability, i.e. $$\forall a>0: \quad P(|X_t|<a)\stackrel{t\rightarrow\infty}{\rightarrow} 0.$$ It is relatively easy to come up with necessary conditions. For example, if $X_t$ diverges in probability, then its moments of even order diverge.

However, the divergence of even-ordered moments is not sufficient. Indeed, a simple counterexample is $X_t=t X_0$, where $$X_0=\begin{cases} 0& \text{with probability 1/2}, \\ 1& \text{with probability 1/4}, \\ -1& \text{with probability 1/4}. \end{cases}$$ For this process, for $k\in\mathbb{N}$ we have $\mathbb{E}[X_t^{2n}]=t^{2n}/2\rightarrow\infty$ as $t\rightarrow\infty$ but at the same time we have $\forall a>0$ that $P(|X_t|<a)\stackrel{t\rightarrow\infty}{\rightarrow} 1/2$.

Are there any standard theorems that give sufficient conditions?


1 Answer 1


It appears you are looking for sufficient conditions in terms of expectations of functions of $X_t$. If so, the following may be offered:

\begin{equation} \text{$|X_t|$ diverges to $\infty$ in probability iff $E\frac1{1+|X_t|}\to0$;} \tag{*} \end{equation} here everywhere the convergence is as $t\to\infty$.

Indeed, $|X_t|$ diverges to $\infty$ in probability iff $Y_t:=1/|X_t|$ converges to $0$ in probability, that is, iff $P(Y_t>b)\to0$ for each real $b>0$. Let $f(y):=\frac y{1+y}$.

Then $I\{y>b\}\le f(y)/f(b)$ for $b>0$ and $y\ge0$, where $I\{\cdot\}$ is the indicator function. So, $P(Y_t>b)\le Ef(Y_t)/f(b)=E\frac1{1+|X_t|}/f(b)\to0$ if $E\frac1{1+|X_t|}\to0$, which proves the "if" part of (*). (In fact, here we use the Markov inequality.)

Using inequality $f(y)\le I\{y>b\}+f(b)$ for $b>0$ and $y\ge0$, we see that \begin{equation} E\frac1{1+|X_t|}= Ef(Y_t)\le P(Y_t>b) +f(b)\to f(b) \end{equation} for each real $b>0$ if $Y_t$ converges to $0$ in probability. So, the "only if" part of (*) follows as well, by letting $b\downarrow0$. (This part also follows by the Lebesgue dominated convergence theorem.)

It should also be clear that, in (*), one can replace $E\frac1{1+|X_t|}$ by $E\frac1{1+X_t^2}$ or $E\exp(-X_t^2/2)$ or, more generally, by $Eg(|X_t|)$, where $g$ is (say) any continuous function which is strictly decreasing to $0$ on $[0,\infty)$.

In particular, this allows one to express the condition that $|X_t|$ diverges to $\infty$ in terms of the characteristic functions $c_t(u)=Ee^{iuX_t}$ of $X_t$ -- say, by writing $e^{-x^2/2}=\int_{-\infty}^\infty e^{iux}\frac1{\sqrt{2\pi}}e^{-u^2/2}\,du$ and $\frac2{1+x^2}=\int_{-\infty}^\infty e^{iux}e^{-|u|}\,du$ for real $x$, whence $E\exp(-X_t^2/2)=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty c_t(u)e^{-u^2/2}\,du$ and $E\frac1{1+X_t^2}=\frac12\,\int_{-\infty}^\infty c_t(u)e^{-|u|}\,du$, and so, \begin{equation} \text{$|X_t|\to\infty$ in probability iff $\int_{-\infty}^\infty c_t(u)e^{-u^2/2}\,du\to0$ iff $\int_{-\infty}^\infty c_t(u)e^{-|u|}\,du\to0$.} \end{equation}

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    $\begingroup$ An excellent example of a perfect answer to somebody asking a question about "finding conditions" without specifying the exact terms in which those conditions should be given: elegant, nontrivial, and (I'm ready to bet $100 on that) totally useless :lol: $\endgroup$
    – fedja
    Commented Nov 21, 2017 at 23:44
  • $\begingroup$ @fedja : Essentially, what the answer does is reduce the condition $EI\{|X_t|<a\}\to0$ for all $a>0$ to $Eg(|X_t|)\to0$ for any one of the functions $g$ described in the last paragraph of the asnwer. How useful is this? I won't take your bet. :-) $\endgroup$ Commented Nov 22, 2017 at 0:58
  • $\begingroup$ @fedja : Actually, I have just added a paragraph showing, as a corollary, necessary and sufficient conditions that might possibly be of use when the distributions of $X_t$ are given by their characteristic functions. $\endgroup$ Commented Nov 22, 2017 at 4:53
  • $\begingroup$ Thanks for this nice answer! The usefulness of the criterion will of course be limited by the ease with which $Eg(|X_t|)$ can be controlled. One could perhaps try to find $g$ that makes this easier, but for the diffusion processes hinted at in my question, most expectations are not available in closed form. I wonder then whether the criterion can perhaps be formulated directly in terms of the stochastic differential equation of the system. $\endgroup$
    – S.Surace
    Commented Nov 22, 2017 at 19:01
  • $\begingroup$ @fedja: Indeed, the terms in which the conditions should be formulated are not entirely clear to me at present. Perhaps a useful guideline is that they should be sufficiently tractable for the types of processes I'm most interested in while being somewhat nontrivial. $\endgroup$
    – S.Surace
    Commented Nov 22, 2017 at 19:09

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